题目

Problem

The times taken, in minutes, to complete a word processing task by 250 employees at a particular company are summarised in the table.

Time taken ( $t$ minutes)	$0\leq t<20$	$20\leq t<40$	$40\leq t<50$	$50\leq t<60$	$60\leq t<100$
Frequency	32	46	96	52	24

(a) Draw a histogram to represent this information.

(4)

From the data, the estimate of the mean time taken by these 250 employees is 43.2 minutes.

(b) Calculate an estimate for the standard deviation of these times.

(3)

题目中文翻译

某公司 250 名员工完成一项文字处理任务所需的时间（单位为分钟）汇总于下表。

用时（ $t$ 分钟）	$0\leq t<20$	$20\leq t<40$	$40\leq t<50$	$50\leq t<60$	$60\leq t<100$
频数	32	46	96	52	24

(a) 画一幅直方图表示这些数据。

(4)

根据这些数据，这 250 名员工用时均值的估计值为 43.2 分钟。

(b) 计算这些用时的标准差的估计值。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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不同组距用频数密度作柱高，表格给出所有组宽与密度。

答题过程

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Time interval (minutes)	Class width	Frequency density
$0\leq t<20$	$20$	$32/20=1.6$
$20\leq t<40$	$20$	$46/20=2.3$
$40\leq t<50$	$10$	$96/10=9.6$
$50\leq t<60$	$10$	$52/10=5.2$
$60\leq t<100$	$40$	$24/40=0.6$

Draw adjoining bars over these intervals with the stated heights. Label the axes “Time (minutes)” and “Frequency density”, using linear scales.

(b)

解法一

思路

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用组中点10、30、45、55、80求二阶矩，均值用题给43.2。

答题过程

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The class midpoints are $10,30,45,55,80$ minutes. Thus

\begin{align*} \sum fm^2=&\,32(10)^2+46(30)^2+96(45)^2\\[4mm] &\,\hspace{2pt}+52(55)^2+24(80)^2\\[4mm] =&\,549900. \end{align*}

Using the given mean, the estimated standard deviation is

\begin{align*} \sigma=&\,\sqrt{\frac{549900}{250}-(43.2)^2}\\[4mm] =&\,\sqrt{333.36}=18.258148\ldots\\[4mm] \approx&\,\boxed{18.3\text{ minutes}}. \end{align*}

解法二

思路

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也可以直接计算各组中点与题给均值43.2的平方离差，再按频数加权求平均。

答题过程

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Using the class midpoints and the given mean,

\begin{align*} \sum f(m-43.2)^2 =&\,32(10-43.2)^2+46(30-43.2)^2\\[4mm] &\,\hspace{2pt}+96(45-43.2)^2+52(55-43.2)^2\\[4mm] &\,\hspace{4pt}+24(80-43.2)^2\\[4mm] =&\,83340. \end{align*}

Therefore the estimated standard deviation is

\sigma=\sqrt{\frac{83340}{250}} =\sqrt{333.36}\approx\boxed{18.3\text{ minutes}}.

CIE 9709 2022 November Paper 52 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程